Problem: Graph this system of equations and solve. $y = \dfrac{1}{3} x - 3$ $-6x-3y = -12$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Explanation: The y-intercept for the first equation is $-3$ , so the first line must pass through the point $(0, -3)$ The slope for the first equation is $\dfrac{1}{3}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $3$ positions to the right. $3$ positions to the right. Graph the blue line so it passes through $(0, -3)$ and $(3, -2)$ Convert the second equation, $-6x-3y = -12$ , to slope-intercept form. $y = -2 x + 4$ The y-intercept for the second equation is $4$ , so the second line must pass through the point $(0, 4)$ The slope for the second equation is $-2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) You must also move $1$ positions to the right. $1$ position to the right. $2$ positions down from $(0, 4)$ is $(1, 2)$ Graph the green line so it passes through $(0, 4)$ and $(1, 2)$ The solution is the point where the two lines intersect. The lines intersect at $(3, -2)$.